1
$\begingroup$

Let $(\mu_{n})$ sequence of probability measures of $\mathbb{R}^{d}$ converging to the prob measure $\mu$. Then by definition we know that $\int f d\mu_{n} \longrightarrow \int f d\mu $ for f continuous and bounded function.

I was wondering if I can write the formula above as $\int\int f(x-y) d\mu_{n}(x)d\mu_n(y) \longrightarrow \int\int f(x-y) d\mu(x)d\mu(y) $ ?

Or is the latter formula implied by the definition of weak convergence ?

Do I need additional assumptions on f ?

$\endgroup$

1 Answer 1

1
$\begingroup$

$\newcommand{\eD}{\overset{\text{D}}\to}$

Let $X_n$ and $X$ be any random vectors with distributions $\mu_n$ and $\mu$, respectively. Let $Y_n$ be an independent copy of $X_n$, for each $n$, and let $Y$ be an independent copy of $X$. The questions can then be restated as follows:

Q1: Does $X_n\eD X$ imply $X_n-Y_n\eD X-Y$?

Q2: Vice versa, does $X_n-Y_n\eD X-Y$ imply $X_n\eD X$?

Here $\eD$ denotes the convergence in distribution.

The answer to Q1 is yes. Indeed, let $g_Z$ denote the characteristic function (c.f.) of a random vector $Z$. Then $X_n\eD X$ means that $g_{X_n}\to g_X$ pointwise, whence $g_{X_n-Y_n}=|g_{X_n}|^2\to|g_X|^2=g_{X-Y}$ pointwise, so that $X_n-Y_n\eD X-Y$.

The answer to Q2 is no. Indeed, let e.g. $X_n=n=Y_n$ and $X=Y=0$. Then $X_n-Y_n\eD X-Y$, but $X_n\eD X$ is false.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.